Answer: you can watch a video on how to solve this question on you tube
Answer: Rupture strength
Explanation: Rupture strength is the strength of a material that is bearable till the point before the breakage by the tensile strength applied on it. This term is mentioned when there is a sort of deformation in the material due to tension.So, rupture will occur before whenever there are chances of failing and the material is still able to bear stresses before failing.
Answer:
4.17x10^-3 kW/K
Explanation:
Detailed explanation and calculation is shown in the image below
Answer:
d= 4.079m ≈ 4.1m
Explanation:
calculate the shaft diameter from the torque, \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}
Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).
r = Radius of the shaft.
T = Twisting Moment or Torque.
J = Polar moment of inertia.
C = Modulus of rigidity for the shaft material.
l = Length of the shaft.
θ = Angle of twist in radians on a length.
Maximum Torque, ζ= τ × \frac{ π}{16} × d³
τ= 60 MPa
ζ= 800 N·m
800 = 60 × \frac{ π}{16} × d³
800= 11.78 × d³
d³= 800 ÷ 11.78
d³= 67.9
d= \sqrt[3]{} 67.9
d= 4.079m ≈ 4.1m
Answer:
#include <iostream>
#include <iomanip>
using namespace std;
class pointType
{
public:
pointType()
{
x=0;
y=0;
}
pointType::pointType(double x,double y)
{
this->x = x;
this->y = y;
}
void pointType::setPoint(double x,double y)
{
this->x=x;
this->y=y;
}
void pointType::print()
{
cout<<"("<<x<<","<<y<<")\n";
}
double pointType::getX()
{return x;
}
double pointType::getY()
{return y;
}
private:
double x,y;
};
int main()
{
pointType p2;
double x,y;
cout<<"Enter an x Coordinate for point ";
cin>>x;
cout<<"Enter an y Coordinate for point ";
cin>>y;
p2.setPoint(x,y);
p2.print();
system("pause");
return 0;
}
